Solved Examples
(iv) 2 same 2 another same Selection
(2X) (2Y) 3C2 = 3
Arrangements = (2+2)!/2!2! = 6
Total words = 3 × 6 = 18
(v) 1 same, 3 diff º 4diff. Selection 6C4 = 6C2 = 15
Arrangement = (1+1+1+1)!/1!1!1!1! = 4! = 24
Total words = 15 × 24 = 360
(vi) 1 same, 3 another same º case ii (already considered)
Hence grand total of desired words
= 20 + 360 + 18 + 360 - 758.
Example 9
You are given the responsibility of organizing a fresher's welcome party at IIT-Delhi. Total 496 students will join IIT-D. Six restaurants A, B, C, D, E and F are booked. Capacity of each is 120, 146, 46, 72, 72, 80 persons at a time respectively. You have to group them in such a way that each group has least possible number of students. How would you do so?
Solution:
Remember: Equal division minimizes the number of students in each group.
But: 496/6 = 82.6 > 46
Therefore C has least capacity of 46 and is not capable to accommodate 82 students. Hence let us group for it first i.e. 46 students get accommodated.
Students left Restaurant left
450 A B D E and F
But 450/5 = 90 > 72
Next D and E each require smallest group 72, 72
D → 72 E → 72
Students left Restaurant left
306 A B and F
Now F is the least left for grouping, so assigning 80 to F.
Students left Restaurant left
226 A and B
Now Therefore 226/2 = 113 < 120 and each of A and B is capable of accommodating to this. Hence,
A → 113
B → 113
But not
A → 120
B → 106
As in the case A does not have minimum possible number of students.
Hence, we grouped 496 students into 46, 72, 72, 80, 113 and 113 each.
No. of ways of doing so = 496! / (46!(72!)2 .2!80!(113!)2.2!)